Mechanical Engineering Principles

SIMPLE MACHINES 201

Effort

(i) (ii)

Load

(b)

Figure 18.1(b)

Effort

(iii) (ii)

(i)

Load

(c)

Figure 18.1(c)

and since from equation (18.3),

efficiency=

force ratio

movement ratio

×100%

it follows that when the force ratio isn,

100 =

n

movement ratio

× 100

that is, the movement ratio is alson.

Problem4. Aloadof80kgisliftedbya

three-pulley system similar to that shown in

Figure 18.1(c) and the applied effort is

392 N. Calculate (a) the force ratio, (b) the

movement ratio, (c) the efficiency of the

system. Takegto be 9.8 m/s^2.

(a) From equation (18.1), the force ratio is given

by

load

effort

The load is 80 kg, i.e. (80× 9 .8) N, hence

force ratio=

80 × 9. 8

392

= 2

(b) From above, for a system havingnpulleys, the

movement ratio isn. Thus for a three-pulley

system, themovement ratio is 3

(c) From equation (18.3),

efficiency=

force ratio

movement ratio

× 100

=

2

3

× 100 = 66 .67%

Problem 5. A pulley system consists of two

blocks, each containing three pulleys and

connected as shown in Figure 18.2. An effort

of 400 N is required to raise a load of

1500 N. Determine (a) the force ratio, (b) the

movement ratio, (c) the efficiency of the

pulley system.

Figure 18.2

(a) From equation (18.1),

force ratio=

load

effort

=

1500

400

= 3. 75